is equivalent with the sum of terms like $ a(\frac{1}{\beta}+\frac{1}{\gamma}-\frac{2}{\alpha}) $ is $ \geq 0 $. now cebisev.

anonymouslonely wrote: is equivalent with the sum of terms like $ a(\frac{1}{\beta}+\frac{1}{\gamma}-\frac{2}{\alpha}) $ is $ \geq 0 $. now cebisev. Are you sure? The terms all have to be positive in the chebyshev inequality which I think you are referring to here.

the numbers in cebisev's inequality don't have to be positive.

thank you - you are right, and my notes are wrong

WakeUp wrote: Let $\alpha,\beta,\gamma$ be the angles of a triangle opposite to its sides with lengths $a,b,c$ respectively. Prove the inequality \[a\left(\frac{1}{\beta}+\frac{1}{\gamma}\right)+b\left(\frac{1}{\gamma}+\frac{1}{\alpha}\right)+c\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)\ge2\left(\frac{a}{\alpha}+\frac{b}{\beta}+\frac{c}{\gamma}\right)\] Kazakstan 2011